WORST_CASE(Omega(1),?) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg2==arg2P_1 && arg3==arg3P_1 ], cost: 1 1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && 0==arg2P_2 && arg3==arg3P_2 ], cost: 1 2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 && 0==arg3P_3 ], cost: 1 5: f4 -> f5 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1>10 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1 6: f4 -> f6 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1<=10 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1 3: f5 -> f8 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1==arg1P_4 && 1==arg2P_4 && arg3==arg3P_4 ], cost: 1 7: f8 -> f7 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1 4: f6 -> f9 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg1==arg1P_5 && arg2==arg2P_5 && 1==arg3P_5 ], cost: 1 8: f9 -> f7 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg1==arg1P_9 && arg2==arg2P_9 && arg3==arg3P_9 ], cost: 1 9: f7 -> f7 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg2==arg3 && arg1==arg1P_10 && arg2==arg2P_10 && arg3==arg3P_10 ], cost: 1 10: f7 -> f10 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg2 f10 : arg1'=arg1P_12, arg2'=arg2P_12, arg3'=arg3P_12, [ arg2>arg3 && arg1==arg1P_12 && arg2==arg2P_12 && arg3==arg3P_12 ], cost: 1 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], cost: 1 Removed unreachable and leaf rules: Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg2==arg2P_1 && arg3==arg3P_1 ], cost: 1 1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && 0==arg2P_2 && arg3==arg3P_2 ], cost: 1 2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 && 0==arg3P_3 ], cost: 1 5: f4 -> f5 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1>10 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1 6: f4 -> f6 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1<=10 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1 3: f5 -> f8 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1==arg1P_4 && 1==arg2P_4 && arg3==arg3P_4 ], cost: 1 7: f8 -> f7 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1 4: f6 -> f9 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg1==arg1P_5 && arg2==arg2P_5 && 1==arg3P_5 ], cost: 1 8: f9 -> f7 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg1==arg1P_9 && arg2==arg2P_9 && arg3==arg3P_9 ], cost: 1 9: f7 -> f7 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg2==arg3 && arg1==arg1P_10 && arg2==arg2P_10 && arg3==arg3P_10 ], cost: 1 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], cost: 1 Simplified all rules, resulting in: Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1 1: f2 -> f3 : arg2'=0, [], cost: 1 2: f3 -> f4 : arg3'=0, [], cost: 1 5: f4 -> f5 : [ arg1>10 ], cost: 1 6: f4 -> f6 : [ arg1<=10 ], cost: 1 3: f5 -> f8 : arg2'=1, [], cost: 1 7: f8 -> f7 : [], cost: 1 4: f6 -> f9 : arg3'=1, [], cost: 1 8: f9 -> f7 : [], cost: 1 9: f7 -> f7 : [ arg2==arg3 ], cost: 1 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 8. Accelerating the following rules: 9: f7 -> f7 : [ arg2==arg3 ], cost: 1 Accelerated rule 9 with non-termination, yielding the new rule 13. [accelerate] Nesting with 0 inner and 0 outer candidates Removing the simple loops: 9. Accelerated all simple loops using metering functions (where possible): Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1 1: f2 -> f3 : arg2'=0, [], cost: 1 2: f3 -> f4 : arg3'=0, [], cost: 1 5: f4 -> f5 : [ arg1>10 ], cost: 1 6: f4 -> f6 : [ arg1<=10 ], cost: 1 3: f5 -> f8 : arg2'=1, [], cost: 1 7: f8 -> f7 : [], cost: 1 4: f6 -> f9 : arg3'=1, [], cost: 1 8: f9 -> f7 : [], cost: 1 13: f7 -> [11] : [ arg2==arg3 ], cost: NONTERM 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], cost: 1 Chained accelerated rules (with incoming rules): Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1 1: f2 -> f3 : arg2'=0, [], cost: 1 2: f3 -> f4 : arg3'=0, [], cost: 1 5: f4 -> f5 : [ arg1>10 ], cost: 1 6: f4 -> f6 : [ arg1<=10 ], cost: 1 3: f5 -> f8 : arg2'=1, [], cost: 1 7: f8 -> f7 : [], cost: 1 14: f8 -> [11] : [ arg2==arg3 ], cost: NONTERM 4: f6 -> f9 : arg3'=1, [], cost: 1 8: f9 -> f7 : [], cost: 1 15: f9 -> [11] : [ arg2==arg3 ], cost: NONTERM 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], cost: 1 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1 1: f2 -> f3 : arg2'=0, [], cost: 1 2: f3 -> f4 : arg3'=0, [], cost: 1 5: f4 -> f5 : [ arg1>10 ], cost: 1 6: f4 -> f6 : [ arg1<=10 ], cost: 1 3: f5 -> f8 : arg2'=1, [], cost: 1 14: f8 -> [11] : [ arg2==arg3 ], cost: NONTERM 4: f6 -> f9 : arg3'=1, [], cost: 1 15: f9 -> [11] : [ arg2==arg3 ], cost: NONTERM 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], cost: 1 Eliminated locations (on linear paths): Start location: __init 21: f4 -> [11] : [ arg1>10 && 1==arg3 ], cost: NONTERM 22: f4 -> [11] : [ arg1<=10 && arg2==1 ], cost: NONTERM 18: __init -> f4 : arg1'=arg1P_1, arg2'=0, arg3'=0, [], cost: 4 Eliminated locations (on tree-shaped paths): Start location: __init ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Constant Cpx degree: 0 Solved cost: 1 Rule cost: 1 Rule guard: [] WORST_CASE(Omega(1),?)